# Why do both sine and cosine exist?

Cosine is just a change in the argument of sine, and vice versa.

So why do we have both of them? Do they both exist simply for convenience in defining the other trig functions?

You seem to be asking why we name them both, rather than why they exist, since the very relationships you’ve written shows that if one exists, the other does, too.

Essentially, all mathematical notation and names, except for a very small subset, are for convenience/clarity/human communication. Math does not require that we name any function consistently across separate proofs, but it becomes much easier to communicate and think about things when we do.

I prefer the relationship:

$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$ since this is symmetric and obviously geometric when $$0\leq x\leq \pi/20\leq x\leq \pi/2$$, and because this is also the relationship between “tangent” and “cotangent” and “secant” and “cosecant.” It indicates a duality in these functions.

It is certainly something of a paradox that adding more names often simplifies our understanding. In particular, if you defined only one, it would give you a sense that one of these functions was “primary.” There is a hint of that error even in the names “sine” and “cosine,” which vaguely implies that “sine” is primary, but it would be particularly strong if we only defined “sine” and never defined “cosine.” We would have a harder time grasping the duality that happens in trig functions.

If you actually must start with one function, most mathematicians wouldn’t start with $$\cos x\cos x$$ or $$\sin x,\sin x,$$ they’d start with the complex-valued function: $$\operatorname{cis}(x)=\cos(x)+i\sin(x).\operatorname{cis}(x)=\cos(x)+i\sin(x).$$ This has the property $$\operatorname{cis}(x)\operatorname{cis}(y)=\operatorname{cis}(x+y),\operatorname{cis}(x)\operatorname{cis}(y)=\operatorname{cis}(x+y),$$ and can also be written as $$e^{ix}.e^{ix}.$$ You can define $$\cos x\cos x$$ and $$\sin x\sin x$$ in terms of $$\operatorname{cis} x:\operatorname{cis} x:$$

$$\cos x=\frac{1}{2}\left(\operatorname{cis}(x)+\operatorname{cis}(-x)\right)\\ \sin x=\frac{1}{2i}\left(\operatorname{cis}(x)-\operatorname{cis}(-x)\right)\cos x=\frac{1}{2}\left(\operatorname{cis}(x)+\operatorname{cis}(-x)\right)\\ \sin x=\frac{1}{2i}\left(\operatorname{cis}(x)-\operatorname{cis}(-x)\right)$$

Addition: I recently heard, in a lecture that covered some of the accomplishments of the early Muslim world, that they acquired knowledge of the sine function from India, and developed the other trigonometric functions. So if you want someone to blame, we can blame them. 🤓